Linear System with Parameter

Here is problem 3 from the Fifth International Internet Mathematical Olympiad for Students. This is an online competition run by the Ariel University Center of Samaria, Israel.

Find all values of the parameter y for which the following system has a solution

(1)	x5 + x2	= yx1
(2)	x1 + x3	= yx2
(3)	x2 + x4	= yx3
(4)	x3 + x5	= yx4
(5)	x4 + x1	= yx5

For each value of y, find all the solutions.

Solution

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Copyright © 1996-2018 Alexander Bogomolny

Find all values of the parameter y for which the following system has a solution

(1)	x5 + x2	= yx1
(2)	x1 + x3	= yx2
(3)	x2 + x4	= yx3
(4)	x3 + x5	= yx4
(5)	x4 + x1	= yx5

For each value of y, find all the solutions.

First of all observe that, for any value of y, the system has a trivial solution: x₁ = x₂ = x₃ = x₄ = x₅ = 0. The task is to find the values of y for which there is a nontrivial solution. Add up all five equations to obtain

(y - 2)(x1 + x2 + x3 + x4 + x5) = 0.

So, either y = 2 or x₁ + x₂ + x₃ + x₄ + x₅ = 0. Clearly the value y = 2 is bound to play a prominent role in the solution. Curiously, there are other y's of no lesser importance.

But, to follow up, assume first that y = 2. The system becomes

(1')	x5 + x2	= 2x1
(2')	x1 + x3	= 2x2
(3')	x2 + x4	= 2x3
(4')	x3 + x5	= 2x4
(5')	x4 + x1	= 2x5

Since the system is homogeneous and we are looking for a non-trivial solution, we may omit the last equation and solve the rest with the Gaussian elimination, getting subsequently,

- 2 1 0 0 1 1 - 2 1 0 0 0 1 - 2 1 0 0 0 1 - 2 1

⇒

- 2 1 0 0 1 0 - 3 2 0 1 0 0 - 4 3 1 0 0 0 - 5 5

By the back substitution, x₅ = x₄ = x₃ = x₂ = x₁, with arbitrary x₁.

Now we may assume y ≠ 2 and x₁ + x₂ + x₃ + x₄ + x₅ = 0. We'll replace (quite arbitrarily) the last equation with this one and employ again the Gaussian elimination.

1 1 1 1 1 -y 1 0 0 1 1 -y 1 0 0 0 1 -y 1 0 0 0 1 -y 1

⇒

1 1 1 1 1 0 y+1 y y y+1 0 -(y+1) 0 -1 -1 0 1 -y 1 0 0 0 1 -y 1

⇒

	1 1 1 1 1 0 1 -y 1 0 0 y+1 y y y+1 0 -(y+1) 0 -1 -1 0 0 1 -y 1	⇒	1 1 1 1 1 0 1 -y 1 0 0 0 y y-1 y 0 0 -y(y+1) y -1 0 0 1 -y 1	⇒

1 1 1 1 1 0 1 -y 1 0 0 0 -y(y+1) y -1 0 0 y y-1 y 0 0 y+1 -1 y+1

⇒

1 1 1 1 1 0 1 -y 1 0 0 0 -y(y+1) y -1 0 0 0 y²+y-1 y²+y-1 0 0 0 0 y²+y-1

If y² + y - 1 ≠ 0, the last equation gives x₅ = 0 and the back substitution leads to a trivial solution which was already accounted for. So let us assume that y² + y - 1 = 0. The quadratic formula gives two distinct solutions

y1,2 = (-1 ± √5) / 2.

With the parameter y equal to one of them, the last equation implies that x₅ can be chosen arbitrarily; and, from the fourth equation, the same is true of x₄.

Since, for the chosen y, -y(y + 1) = -1, the third equation gives

x3 = yx4 - x5

From the second equation,

x2 = -y(x4 + x5)

and, from the first equation,

x1 = -x4 + yx5.

Thus the system has three sets of solutions:

1. A trivial solution for any y,

2. A 1-parameter solution for y = 2,

3. A 2-parameter solution for y = (-1 ± √5) / 2.

The latter family, perhaps surprisingly, includes the golden ratio.

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Copyright © 1996-2018 Alexander Bogomolny